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I'm trying the following problem:

Verify whether the given binary relation $R$ on the set $A$ is an order relation:

Let $A$ be the set whose elements are the sequences $\{x_n\}_{n\in \mathbb{N}}$ of real numbers, and only these, and let $R$ be the relation given by

$$R = \{(\{x_n\}_{n\in \mathbb{N}}, \{y_n\}_{n\in \mathbb{N}}) \in A \times A: [x_{33} \le y_{33}]\}$$

I don't understand exactly what is meant by the term "order relation". Is it the same as asking if the relation is a partial order (i.e. asking if the relation demonstrates reflexivity, symmetry and transitivity)?

If so, how could I go about showing that in this case? Is the subscript $33$ in the definition referring to the 33rd element of the sequences $x$ and $y$?

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  • $\begingroup$ The term order relation is a bit ambiguous, but here it most likely does mean partial order. And yes, $x_{33}$ is the term of the $x$ sequence with subscript $33$; it’s either the $33$-rd term, if your $\Bbb N$ is really $\Bbb Z^+$, or the $34$-th term, if $0\in\Bbb N$. $\endgroup$ – Brian M. Scott Nov 6 '16 at 20:11
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Yes, the subscript $33$ probably refers to the $33$rd element (if $\mathbb{N} = \{1,2, \dots\}$, if however $\mathbb{N} = \{0,1,2,\dots\}$ it is the 34th element) in the sequence, and an order relation is indeed just $R$ being a partial order. However, you probably mean that a partial order is antisymmetric.

Note that the antisymmetry is going to cause problems here: if $(x,y) \in R$ and $(y,x) \in R$, this tells you that $x_{33} = y_{33}$. But if antisymmetry is to hold, this needs to imply that $x_n = y_n$ for all other $n$ too.

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  • $\begingroup$ Yep, I meant "antisymmetric", no idea why I wrote "symmetric" instead. That clears things up, thanks! $\endgroup$ – enharmonics Nov 6 '16 at 21:12

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